sqrt{2+sqrt{2+sqrt{2+2 cos 8 theta}}} = | vedclass

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(B) Let $y = \sqrt{2+\sqrt{2+\sqrt{2+2 \cos 8 	heta}}}$.We know that $1 + \cos 2A = 2 \cos^2 A$.Using this identity,we simplify the expression step b

Vedclass · Accepted Answer

$2 \cos 	heta$

Vedclass · Answer

(B) Let $y = \sqrt{2+\sqrt{2+\sqrt{2+2 \cos 8 	heta}}}$.We know that $1 + \cos 2A = 2 \cos^2 A$.Using this identity,we simplify the expression step by step:$y = \sqrt{2+\sqrt{2+\sqrt{2(1 + \cos 8 	heta)}}}$$y = \sqrt{2+\sqrt{2+\sqrt{2(2 \cos^2 4 	heta)}}}$$y = \sqrt{2+\sqrt{2+\sqrt{4 \cos^2 4 	heta}}}$$y = \sqrt{2+\sqrt{2+2 \cos 4 	heta}}$$y = \sqrt{2+\sqrt{2(1 + \cos 4 	heta)}}$$y = \sqrt{2+\sqrt{2(2 \cos^2 2 	heta)}}$$y = \sqrt{2+\sqrt{4 \cos^2 2 	heta}}$$y = \sqrt{2+2 \cos 2 	heta}$$y = \sqrt{2(1 + \cos 2 	heta)}$$y = \sqrt{2(2 \cos^2 	heta)}$$y = \sqrt{4 \cos^2 	heta} = 2 \cos 	heta$.

$\sqrt{2+\sqrt{2+\sqrt{2+2 \cos 8 \theta}}} = $

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